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3 years ago

6

When velocity is positive and acceleration is negative, what happens to the object’s motion?

1 answer:

xenn [34]3 years ago

4 0

Answer:

Option (D) : The object slows down.

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Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

4 0

4 years ago

Work is done if you carry a plant across a room at constant velocity. True False

BlackZzzverrR [31]

it is false hope this helps

5 0

4 years ago

Read 2 more answers

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston i

AVprozaik [17]

Answer:

$D = 55.2 cm$

Explanation:

As we know that the total mass of the all four players is given as

$M = 4\times 110$

$M = 440 kg$

diameter of the piston of cheer leader is given as

$d_1 = 16 cm$

are of cross-section is given as

$A_1 = \pi r^2$

$A_1 = \pi(0.08)^2 = 0.02 m^2$

mass of the cheer leader is given as

m = 55 kg

so the pressure due to cheer leader is given as

$P_{in} = \frac{mg}{A_1}$

$P_{in} = \frac{55 \times 9.81}{0.02}$

$P_{in} = 26835 Pa$

Now on the other side pressure must be same

so we have

$\frac{Mg}{A} + \rho gH = P_{in}$

$\frac{440 \times 9.8}{A} + (900)(9.8)(1) = 26835$

$A = 0.24 m^2$

$\pi r^2 = 0.24$

$r = 0.276 m$

so diameter on the other side is given as

$D = 2 r$

$D = 55.2 cm$

8 0

3 years ago

Which only list metalloids

HACTEHA [7]

Answer: I think the answer <em><u>MIGHT</u></em> be: Boron, germanium, and tellurium....

Explanation:I know this cuz I had this question on my unit test and I had to look it up online....Hope this helps

I really hope you guys find this helpful and thx so much for rating :D

pls say something in chat if you found this helpful

UwU ;-;

7 0

3 years ago

If your good in science please help me with this question.

Dahasolnce [82]

Answer:

A inertia

B force and acceleration

C force

D action reaction

E action reaction

F

5 0

3 years ago